Question 47 of 70intermediate🔧 ApplyShort Answer1 mark

Find the derivative of the following functions (it is to be understood that a,b,c,d,p,q,ra, b, c, d, p, q, r and ss are fixed non-zero constants and mm and nn are integers): 1ax2+bx+c\frac{1}{ax^2 + bx + c}

Correct Answer

−2ax+b(ax2+bx+c)2-\frac{2ax + b}{(ax^2 + bx + c)^2}

Exercise: Miscellaneous Exercise on Chapter 12 | Q: 7 | (Page Page 37)
For More Understanding

Explanation

This question is from the Miscellaneous Exercise on Chapter 12 (Limits and Derivatives) of the NCERT textbook. The question asks to find the derivative of the function where a, b, c are fixed non-zero constants. The solution requires applying the chain rule to differentiate a composite function of the form (denominator)^(-1).

Solution Steps

  1. Step 1: Let f(x)=1ax2+bx+c=(ax2+bx+c)−1f(x) = \frac{1}{ax^2 + bx + c} = (ax^2 + bx + c)^{-1}

  2. Step 2: Apply chain rule: f′(x)=−1⋅(ax2+bx+c)−2⋅ddx(ax2+bx+c)f'(x) = -1 \cdot (ax^2 + bx + c)^{-2} \cdot \frac{d}{dx}(ax^2 + bx + c)

  3. Step 3: Differentiate the denominator: ddx(ax2+bx+c)=2ax+b\frac{d}{dx}(ax^2 + bx + c) = 2ax + b

  4. Step 4: Therefore, f′(x)=−1(ax2+bx+c)2⋅(2ax+b)=−2ax+b(ax2+bx+c)2f'(x) = -\frac{1}{(ax^2 + bx + c)^2} \cdot (2ax + b) = -\frac{2ax + b}{(ax^2 + bx + c)^2}

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